Question

(a) Show that <x > is not a maximal ideal in z[x].

(b)List all the subgroups of Z18, along with 3 their generators.

(c)Let H=< (1 2) > and k = < (1 2 3) > be subgroups of S3. Show that S3 = Hk. Is S3 an internal direct product of H and k ? Justify your answer.

(d)Check whether or not { (2, 5), (1, 3), (5, 2), (3, 1) is an equivalence relation on { 1, 2, 3, 5 }.

Answer 1

Answer on Question #47342 – Math – Abstract Algebra

(a) Show that $<x>$ is not a maximal ideal in $z[x]$.

(b) List all the subgroups of $Z18$, along with 3 their generators.

(c) Let $H = \langle (1, 2) \rangle$ and $k = \langle (1, 2, 3) \rangle$ be subgroups of $S_3$. Show that $S_3 = Hk$. Is $S_3$ an internal direct product of $H$ and $k$? Justify your answer.

(d) Check whether or not $\{(2, 5), (1, 3), (5, 2), (3, 1)\}$ is an equivalence relation on $\{1, 2, 3, 5\}$.

Solution:

We know that $[x]$ is just all polynomials with even coefficients. This is not maximal because $Z[x]$ contains many ideals more than $Z$. For example, it contains the ideal $\langle x \rangle$, which is all things that have positive degree (i.e non-constants).

In the case of $Z[x]$, the ideal equal to

$$(x) = \{a_1 x + a_2 x^2 + \dots a^n x^n; \, x_k \in Z, \, 1 \leq k \leq n\}$$

Is a principal ideal of $Z[x]$ generated $x$. $\frac{Z[x]}{(x)} \cong Z$, an integral domain. So, $(x)$ is a prime ideal of $Z[x]$. Let $J = (x, 2)$ the ideal generated by $x$ and $2$ in $Z[x]$. $J$ is a maximal ideal (of $Z[x]$), which contain polynomials of the form:

$$a_0 + a_1 x + a_2 x^2 + \dots a^n x^n; \, a_i \in Z, \, 0 \leq i \leq n$$

Where $a_0$ is even. As $\frac{Z[x]}{J}$ is a field having two elements, $J$ is a maximum ideal of $Z[x]$ containing the prime ideal $(x)$. So, the prime ideal $(x)$ is not a maximum ideal of $Z[x]$. Thus, $Z[x]$ is not a Dedekind domain. $Z[x]$ is not a PID also, as a principal ideal domain has to be a Dedekind domain. This conclusion is also obvious from the fact that $J$ is not a principal ideal of $Z[x]$.

(b) List all the subgroups of $Z_{18}$, along with 3 their generators.

The divisors of 18 are 1, 2, 3, 6, 9, and 12. There is a subgroup of each of these orders, and they are generated by [0], [9], [6], [3], [2], and [1] respectively.

That is, the subgroups of $Z_{18}$ $\langle [0] \rangle$, $\langle [9] \rangle$, $\langle [6] \rangle$, $\langle [3] \rangle$, $\langle [2] \rangle$ and $\langle [1] \rangle$.

(c) Let $H = \langle (1, 2) \rangle$ and $k = \langle (1, 2, 3) \rangle$ be subgroups of $S_3$. Show that $S_3 = Hk$. Is $S_3$ an internal direct product of $H$ and $k$? Justify your answer.

We write $e$ for the identity element of $S_3$, note that $H = \{e, (1,2)\}$ and $K = \{e, (1,2,3), (1,3,2)\}$. So $Hk$ clearly contains $e, (1,2), (1,2,3)$, and $(1,3,2)$. The computations</x>

$$\begin{array}{l} (1,2)(1,2,3) = (2,3) \\ (1,2)(1,3,2) = (1,3) \end{array}$$

We need to show that HK also contains (2,3) and (1,3). Since $S_{3} = \{\mathrm{e}, (1,2), (1,3), (2,3), (1,2,3), (1,3,2)\}$ we have shown that Hk contains every element of $S_{3}$, and hence $S_{3} = \mathsf{Hk}$.

We note that H and k are both abelian groups (because e.g. they are both cyclic). A direct product of abelian groups (whether internal or external) is abelian. Since $S_{3}$ is not abelian, this tells us that $S_{3}$ is not an internal direct product of H and K. (We know that any group of order less than 6 is abelian, this argument shows more: $S_{3}$ is not an internal direct product of any two of its proper subgroups.)

(d) Check whether or not $\{(2,5), (1,3), (5,2), (3,1)\}$ is an equivalence relation on $\{1,2,3,5\}$.

In mathematics, an equivalence relation is the relation that holds between two elements if and only if they are members of the same cell within a set that has been partitioned into cells such that every element of the set is a member of one and only one cell of the partition. The intersection of any two different cells is empty; the union of all the cells equals the original set. These cells are formally called equivalence classes.

A relation R on a set A is an equivalence relation if and only if R is

- reflexive,

- symmetric, and

- transitive.

Let R be a relation on set A. Any equivalence relation must be reflexive i.e. for all a in A we must have (a, a) in R.

In our example we must have all (1, 1), (2, 2), (3, 3), (5, 5) present in R which we don't. In fact, none are in R.

Therefore Is not an equivalence relation, R is not reflexive, since (1, 1)∉ R for example.

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